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Zorluk: KolayParallel Circuits

A 36 V36\text{ V} direct-current power supply is connected across a parallel circuit containing two branches. If the first branch contains an 18 Ω18\ \Omega resistor and the second branch contains a 9 Ω9\ \Omega resistor, what is the current flowing through the second branch in amperes?

Cevap: 4 A

Cevap

4 A
In a parallel circuit, the voltage across all individual branches is identical to the supply voltage (36 V36\text{ V}). Using Ohm's law for the second branch, the current is calculated as I=VR=36 V9 Ω=4 AI = \frac{V}{R} = \frac{36\text{ V}}{9\ \Omega} = 4\text{ A}.

Adım Adım Çözüm

1
Determine the voltage drop across the second branch.
The voltage across the second branch is 36 V36\text{ V}.
In a parallel circuit, each individual branch experiences the full source voltage (Vsource=V1=V2V_{\text{source}} = V_1 = V_2).
2
Calculate the branch current using Ohm's law.
I2=36 V9 Ω=4 AI_2 = \frac{36\text{ V}}{9\ \Omega} = 4\text{ A}.
Current through an individual resistor branch is found by dividing the branch voltage by the branch resistance (I=VRI = \frac{V}{R}).

Anahtar Kavram

In a parallel circuit, the voltage across every parallel branch is equal to the source voltage, and the current through any branch depends solely on that branch's resistance via Ohm's law (I=VRI = \frac{V}{R}).
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